Rock Launching from a catapult - 2D Kinematics

AI Thread Summary
A catapult launches a rock from a height of 33.0 m at an unknown angle θ and speed v0, with the rock remaining in flight for 6.00 seconds and traveling a horizontal distance of 141.0 m. To solve for θ, v0, and the maximum height, the motion can be analyzed using projectile motion equations, separating the horizontal and vertical components. The horizontal distance allows for the calculation of v_x, while the vertical motion can be analyzed using the initial height and the time of flight to find v_y. Once both components are determined, they can be used to derive the angle and initial speed, and the maximum height can be calculated using the derived values. The discussion emphasizes the importance of resolving the motion into x and y components for accurate calculations.
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Homework Statement



A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H of 33.0 m above sea level, directed at an angle θ above the horizontal with an unknown speed v0

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The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 141.0 m. Assuming that air friction can be neglected, calculate the value of the angle θ.
Calculate the speed at which the rock is launched.
To what height above sea level does the rock rise?

Homework Equations



v = v0 +at
x-x0+ v0t + 1/2at^2
v^2 = v0^2 + 2a(x-x0)
x-x0 = 1/2(v0+v)t


The Attempt at a Solution



To be honest I really had no idea how to do this one, all I know is that you have to find theta first before you can do anything else I couldn't find that, so I couldn't get anything else. Help would be appreciated.
 
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Well first of all, this is a projectile motion problem. Remember that in such problems, you can resolve the motion into x and y components.

If you need a little bit of refresher, first read this for conceptual understanding

http://www.physicsclassroom.com/Class/vectors/U3L2a.html

Then this for quantitative analysis with java applet

http://www.ngsir.netfirms.com/englishhtm/ThrowABall.htm

At any rate, the equations you should use

v_x = v_0\cos(\theta)
v_y = v_0\sin(\theta) - gt
s_x = s_x_0 + v_xt
s_y = s_y_0 + v_yt - 0.5gt^2

You have two unknowns to calculate for : the initial velocity, the angle which the rock was launched at and the maximum height of the rock. You can easily find the last part, if you are able to find the initial velocity and the angle, so at the present moment, you have two unknowns.

You know the projectile remains in flight for 6 seconds. When the projectile lands, you know the y distance is equal to 0. Plug in 0 into the s_y equation, solve for v_y. You now have one equation. Note that you have an initial height.

Next, you know the projectile travels in horizontal direction of 141.0m during the time of flight. Plug 141.0m into s_x equation, solve for v_x. Note that your initial x distance is 0 in this case.

You now have two equations and two unknowns. Solve for the initial velocity and the angle.

Now as for finding the maximum height, if you know calculus, you can derive the s_y equation, set it equal to 0, find the time which the projectile is at the highest time and then plug that time into your s_y equation.

If not, then you can do this the other way. When does the object reach its maximum height? Throw a rock vertically, and what do you know about the rock when its at its maximum height? Specifically speaking, what is the velocity equal to at that particular point?

Use the same information for this problem. What is the y-component of velocity equal to when the projectile reaches the maximum height? Plug that particular velocity into your v_y equation, solve for time, then plug that particular time into s_y equation.
 
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For clarification is my s initial in the y direction the 33m?
 
delecticious said:
For clarification is my s initial in the y direction the 33m?

Correct.
 
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